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a spring is stretched 6 cm when a mass of 200 g is hung on it. calculate the spring constant of this spring.

User Lfkwtz
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1 Answer

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Answer:

k = 32.7 N/m

Step-by-step explanation:

Hooke's Law:

F=−kx

Where:

F is the force applied to the spring (N).

k is the spring constant (N/m).

x is the displacement of the spring from its equilibrium position (m).

x = 6 cm = 0.06

mass = m = 200 g = 0.2 kg

The force applied to the spring is equal to the weight of the mass,

F=m⋅g

g is the acceleration due to gravity = 9.81 m/s²

F = mg = (0.2kg)(9.81m/s²) = 1.962 N

F=−kx

-k = F/x = (1.962 N) / (-0.06 m ) = -32.7 N/m

k = 32.7 N/m

User Lagot
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